Magnetic Dipole: Field, Torque and Energy

Derivation

Magnetic Dipole: Field, Torque and Energy

D-054 Home PU-102 Threads force · energy · fields Depends on Magnetic Vector Potential and Gauge Freedom, Multipole Expansion of the Potential

Statement

For a steady, spatially localized current distribution J(r′) the leading non-vanishing term of the vector potential is the magnetic dipole term, characterized by a single vector m = ½ ∫ r′ × J d³r′ (for a planar loop, m = I a n̂). We derive the far vector potential A = (μ₀/4π)(m × r̂)/r², the corresponding field B = (μ₀/4π)[3(m·r̂)r̂ − m]/r³, the torque N = m × B on the moment in a uniform field, and the orientation energy U = −m·B.

Why it matters

The magnetic moment is the single number that survives when a complicated current loop is viewed from far away: atoms, nuclei, bar magnets, MRI spins, planetary dynamos and antenna coils are all described, to leading order, by m alone. It is the magnetic counterpart of the electric dipole moment, and the torque and energy expressions are what make a compass point north, an NMR spin precess, and a motor turn.

Because the magnetic monopole term vanishes identically for steady currents, the dipole is the leading multipole — there is no 1/r tail as there is in electrostatics. That single fact controls the entire structure of magnetostatic fields far from their sources.

Assumptions

Steady currents (∇·J = 0).Without this the current-conservation identities used to kill the monopole term and to symmetrize the first moment fail; a time-varying source radiates and the static multipole expansion is no longer valid. Localized source, field point outside it (r ≫ source size).The Taylor expansion of 1/|r − r′| in powers of r′/r diverges inside or near the distribution, so the dipole form is only the far-field limit. Uniform external B for the torque and energy.If B varies over the loop there is also a net force F = ∇(m·B) and the torque acquires higher-moment corrections; N = m × B holds only to leading order about the loop centre. Rigid moment of fixed magnitude.If |m| depends on orientation or field (induced/diamagnetic response), the energy is no longer simply −m·B and a factor-of-two subtlety appears for field-induced moments.

Derivation

A(r) = (μ₀/4π) ∫ J(r′) / |r − r′| d³r′

Magnetostatic vector potential in Coulomb gauge, taken as a prior result. A

1/|r − r′| = 1/r + (r̂·r′)/r² + O(r′²/r³)

Multipole (Taylor) expansion for r ≫ r′; the coefficients are Legendre polynomials in cos γ = r̂·r̂′. A

∫ J d³r′ = 0 ⇒ monopole term vanishes

For steady currents ∫ Jᵢ d³r = ∫ ∇·(xᵢ J) d³r = ∮ xᵢ J·da = 0, using ∇·J = 0 and J = 0 on a bounding surface. There is no magnetic monopole. C

A(r) = (μ₀ / 4πr²) ∫ (r̂·r′) J(r′) d³r′

Keep the first surviving (dipole) term; r̂ and r are constants of the integration. A

∫ (x′ᵢ Jⱼ + x′ⱼ Jᵢ) d³r′ = 0

Lemma from ∇·J = 0: ∫ ∇·(xᵢxⱼ J) d³r = 0 gives ∫ (xᵢJⱼ + xⱼJᵢ) d³r = 0, so the symmetric part of the first current moment vanishes and only the antisymmetric part survives. C

∫ (r̂·r′) J d³r′ = −½ r̂ × ∫ (r′ × J) d³r′

Contract the antisymmetric moment: using [r̂×(r′×J)]ᵢ = x′ᵢ(r̂·J) − Jᵢ(r̂·r′) and the Step-5 identity, the mixed term reorganizes into this cross product. C

m ≡ ½ ∫ r′ × J(r′) d³r′ ⟶ A(r) = (μ₀/4π) (m × r̂) / r²

Define the magnetic dipole moment; substituting Step 6 into Step 4 gives the dipole vector potential. For a filamentary loop J d³r′ → I dl, so m = (I/2)∮ r′×dl = I a n̂. B

B = ∇ × A = (μ₀/4π) [ 3(m·r̂)r̂ − m ] / r³

Curl of (m×r)/r³ using ∇(1/r³) and ∇·(r/r³) = 4π δ³(r); the delta term is dropped for the far field r ≠ 0. This is the dipole field. C

N = ∫ r′ × (J × B) d³r′ = m × B

Torque about the centre for uniform B. Expanding r′×(J×B) = J(r′·B) − B(r′·J) and applying the Step-5 symmetry identity to each Cartesian component collapses the integral to (½∫r′×J)×B = m×B. C

U = −∫ N dθ = −∫ mB sinθ dθ = −m·B

Work done against the torque in rotating the rigid moment from θ = π/2 to θ; the constant of integration is fixed by U(90°) = 0. B

Result

m = ½ ∫ r′ × J d³r′ = I a n̂

B(r) = (μ₀/4π) [3(m·r̂)r̂ − m] / r³

N = m × B , U = −m·B

Reading. A current loop looks, from far away, like a point dipole of moment m whose field falls as 1/r³ and has the same angular shape as an electric dipole's. Placed in an external field, the dipole feels a torque m×B that tries to align it with B, and its orientation energy is lowest (U = −mB) when aligned, highest (+mB) when anti-aligned.

Units check. [m] = A·m². Field: [μ₀/4π][m]/[r³] = (T·m/A)(A·m²)/m³ = T. ✓ Torque: [m][B] = (A·m²)(T) = A·m²·kg·A⁻¹·s⁻² = kg·m²·s⁻² = N·m. ✓ Energy: same product → J. ✓

Limiting cases

On axis (r̂ ∥ m): B = (μ₀/4π)(2m/r³), pointing along m — the strongest direction.
Equatorial plane (r̂ ⊥ m): B = −(μ₀/4π)(m/r³), i.e. anti-parallel to m and half the axial magnitude.
Aligned dipole (θ → 0): N → 0, U → −mB (stable equilibrium).
Perpendicular (θ = 90°): torque is maximal, N = mB, while U = 0.
Small oscillations about alignment: U ≈ −mB + ½mBθ², a harmonic well ⇒ a compass/needle librates like a torsional pendulum.

Breaks when

Time-varying currents. If ∂ρ/∂t ≠ 0 then ∇·J ≠ 0, the identities of Steps 3, 5 and 9 collapse, the source radiates, and the static 1/r³ dipole field is replaced by retarded radiation fields (1/r at large distance).
Near or inside the source. For r comparable to the loop size the multipole series does not converge; the true field includes quadrupole and higher terms, and on the loop axis close in one must use Biot–Savart directly.
Strongly non-uniform external field. N = m×B and U = −m·B assume B constant over the source; a gradient adds a translational force F = ∇(m·B) and reference-point ambiguity in the torque.
Field-induced or saturating moments. If m is itself a response to B (diamagnets, superconductors, ferromagnet saturation), U = −m·B is wrong — induced moments carry an extra ½ and the constitutive relation must be tracked.

Failure modes

Dropping the ½. Writing m = ∫ r′×J instead of ½∫ — the factor comes from symmetrizing the first moment (Step 5), and forgetting it doubles every predicted field and torque.
Using the electric-dipole field factor. The angular structure 3(m·r̂)r̂ − m is identical to the electric case, but students sometimes insert 1/(4πε₀) instead of μ₀/4π.
Sign of the equatorial field. Assuming B points along m everywhere; in the equatorial plane it is anti-parallel.
Confusing torque angle and energy angle. N ∝ sinθ but U ∝ −cosθ; using sinθ in the energy (or cosθ in the torque) is a frequent slip.
Applying U = −m·B to the total energy. It is only the orientation energy in an external field; it omits the self-energy and the work done by the source keeping I constant.
Forgetting N = m×B requires uniform B. Applying it in a gradient and ignoring the net force.

Discussion

The deep reason the magnetic multipole expansion starts at the dipole is charge conservation: ∇·J = 0 forbids a net "magnetic charge," so the 1/r monopole potential that dominates electrostatics simply does not exist here. Everything magnetic that we see from a distance — the Earth's field, a fridge magnet, an electron — is dipolar to leading order, and the vector m is the compact carrier of that information.

The torque and energy expressions unify a remarkable range of physics. A compass needle is a permanent moment seeking the −m·B minimum; an electric motor exploits N = m×B with a commutator to keep the torque one-signed; NMR and MRI rest on the fact that a moment not aligned with B feels a torque perpendicular to both m and B, which for a moment tied to angular momentum (m = γL) produces steady Larmor precession rather than alignment.

The gyromagnetic link m = γL is where classical and quantum descriptions meet. Classically a rotating charge gives γ = q/2M, so orbital moments come in units of the Bohr magneton μ_B = eℏ/2m_e. Spin, however, carries an anomalous factor g ≈ 2 (from the Dirac equation, corrected by QED to g = 2.00232…), so the same torque law N = m×B governs both, but with a magnitude no classical loop model can reproduce. The precession frequency ω = γB is what a spectrometer actually measures.

Common misconceptions. The dipole field formula is a far-field result; it is not the field at the centre of a real loop (that is μ₀I/2R, finite, not divergent). Also, m×B tends to align a static moment with B, but a moment carrying angular momentum precesses instead of aligning — alignment requires dissipation. Finally, "the loop moves to strong field" is only true for a moment already aligned; an anti-aligned moment is pushed toward weak field.

Worked examples

Far axial field of a small current loop.

A single circular loop of radius R = 5.0 cm carries I = 2.0 A. Find its moment and the field on the axis at z = 0.50 m. A

m = I a = I πR² = (2.0)(π)(0.050)² = 1.571 × 10⁻² A·m²

Planar loop, a = πR². A

B_axis = (μ₀/4π)(2m/z³) = (10⁻⁷)(2 × 1.571×10⁻²)/(0.50)³

Axial case r̂ ∥ m, and z = 0.50 m ≫ R so the dipole limit is valid. B

B_axis = (10⁻⁷)(3.142×10⁻²)/(0.125) = 2.51 × 10⁻⁸ T ≈ 25 nT

Reading. At ten loop-radii the field is only tens of nanotesla — about the Earth's field divided by a thousand — and directed along m.

Units check. (T·m/A)(A·m²)/m³ = T. ✓

Torque and energy of the same loop in a uniform field.

The loop (m = 1.571×10⁻² A·m²) sits in a uniform B = 0.30 T with m at θ = 30° to B. Find the torque magnitude and the orientation energy. A

N = mB sinθ = (1.571×10⁻²)(0.30)(sin 30°) = (1.571×10⁻²)(0.30)(0.500)

Magnitude of m×B. A

U = −mB cosθ = −(1.571×10⁻²)(0.30)(cos 30°) = −(1.571×10⁻²)(0.30)(0.866)

Orientation energy relative to the θ = 90° reference. A

N = 2.36 × 10⁻³ N·m , U = −4.08 × 10⁻³ J

Reading. The torque drives the loop toward alignment; the energy is negative because θ < 90° is on the favourable side. To flip it fully to θ = 180° would cost U(180°) − U(30°) = (+4.71 − (−4.08))×10⁻³ = 8.79 × 10⁻³ J.

Units check. (A·m²)(T) = N·m for torque and J for energy. ✓

Problems

A flat coil of N = 50 turns and area a = 1.2 × 10⁻³ m² carries I = 0.40 A. Find its magnetic moment.
Solution
m = NIa = (50)(0.40)(1.2×10⁻³) = 2.4 × 10⁻² A·m². The turns add coherently, so an N-turn coil has N times the single-loop moment.
For a point dipole m, find the ratio of the field magnitude on the axis to that in the equatorial plane at the same distance r, and state the direction in each case.
Solution
Axial: B_ax = (μ₀/4π)(2m/r³), along +m. Equatorial: B_eq = (μ₀/4π)(m/r³), along −m. Ratio |B_ax|/|B_eq| = 2. The axial field is twice as strong and points opposite to the equatorial field.
The loop of Worked Example 1 (m = 1.571×10⁻² A·m²) is placed in B = 0.50 T. Find (a) the maximum torque and (b) the work needed to rotate it from fully aligned to fully anti-aligned.
Solution
(a) N_max = mB = (1.571×10⁻²)(0.50) = 7.85 × 10⁻³ N·m, occurring at θ = 90°. (b) W = U(180°) − U(0°) = (+mB) − (−mB) = 2mB = 2(7.85×10⁻³) = 1.57 × 10⁻² J.
A compass needle of magnetic moment m = 0.80 A·m² and moment of inertia I_r = 3.0 × 10⁻⁶ kg·m² oscillates in a horizontal field B = 2.0 × 10⁻⁵ T. Find the period of small oscillations.
Solution
Equation of motion I_r θ̈ = −mB sinθ ≈ −mB θ, so ω = √(mB/I_r) = √[(0.80)(2.0×10⁻⁵)/(3.0×10⁻⁶)] = √(5.33) = 2.31 rad/s. Period T = 2π/ω = 2.72 s. This is exactly the torsional-pendulum analogue with mB playing the role of a torsion constant.
Show that a classical particle of charge q and mass M in a circular orbit has m = (q/2M) L, and evaluate the moment for an electron with orbital angular momentum L = ℏ.
Solution
Orbital current I = q/T = qv/(2πR); moment m = IπR² = qvR/2. Angular momentum L = MvR, so vR = L/M and m = (q/2M)L; as vectors m = (q/2M)L. For an electron (q = −e) with L = ℏ: |m| = eℏ/2m_e = μ_B = (1.602×10⁻¹⁹)(1.055×10⁻³⁴)/(2 × 9.109×10⁻³¹) = 9.27 × 10⁻²⁴ A·m², the Bohr magneton. The moment is anti-parallel to L because the charge is negative.